Question

O 66. How many ml of KMnO, solution containing 158 g/L
must be used to complete the conversion of 75 g of KI
to I, in the acidic solution ?
2KMnO4 + 10KI + 8H,SO4 →6K,SO4 +
2MnSO4 + 51, +8H,O​

Answer 1

Explanation:

NKMnO4=31.6×1158=5.0

(Mn7++5e→Mn2+∴E=5M=51580=31.6)

Meq. of KI=16675×1000=451.8(2I−→I2+2e∴E=1M=1166)

Now Meq. of KMnO4=Meq. of KI

5×V=451.8

∴V=90.36mL